IR Transmitter

[gramo]

I'm building a simple IR transmitter, just wanted to know if this circuit is ok, or if it could be improved.

The IR LED has a 1.6V Forward voltage, and operates at a nominal 50mA. The closest I could get, with the below circuit, was 44.9mA, is there a way I could run it right on 50mA?

 

[ericgibbs]

Morning gramo,

If you have the space, you could parallel a 'higher' value across the 68R.

I take it you dont want to use a 'special resistor', how many such LED drivers are you building?

Is the transistor saturated, it appears from the calculations, there is about a Vces of 0.25V,
you could drive it harder with a lower value base resistor.

Regards.

 

[ronsimpson]

Did you measure the current at 44.9mA? There is about 0.3 volts drop across the transistor that you may not have counted on. If you want to bump up the current by 10% drop the resistance by 10% or choose the next smaller value. Only 10% will not effect the range of you IR project.

What duty cycle will you have the LED on. I do not know your application but if you can guarantee that the LED will be on for no more that 50% of the time you can almost double the current. There should be graphs for that LED showing duty cycle and current.

 

[audioguru]

The base current is too low for the transistor to saturate properly.

Every transistor's datasheet shows its saturation voltage with the base current at 1/10th of the collector current.
Your base current is only 0.5mA so the transistor would saturate well if its load current is only 5ma.

A PIC can supply an output current of up to 25mA so try a 680 ohms base resistor for a base current of about 5mA.

 

[gramo]

Morning gramo,

If you have the space, you could parallel a 'higher' value across the 68R.

I take it you don’t want to use a 'special resistor', how many such LED drivers are you building?

Is the transistor saturated, it appears from the calculations, there is about a Vces of 0.25V,
you could drive it harder with a lower value base resistor.

Regards.

I'm about too hit the hay sack, but I'll fit in one more post

I cant take the base resistor lower, but how low can I take Vce (how hard can I drive the transistor) to keep everything safe for long term use?

I just noticed I used an 8.6K resistor on the base - it should be a 8.2K
It doesn't change much, but yeah, you get the picture

 

[gramo]

The base current is too low for the transistor to saturate properly.

Every transistor's datasheet shows its saturation voltage with the base current at 1/10th of the collector current.
Your base current is only 0.5mA so the transistor would saturate well if its load current is only 5ma.

A PIC can supply an output current of up to 25mA so try a 680 ohms base resistor for a base current of about 5mA.

Sorry audioguru, I posted when you did...

Thanks, I'll give it a wherl

 

[Nigel Goodwin]

I'm building a simple IR transmitter, just wanted to know if this circuit is ok, or if it could be improved.

The IR LED has a 1.6V Forward voltage, and operates at a nominal 50mA. The closest I could get, with the below circuit, was 44.9mA, is there a way I could run it right on 50mA?

Instead of wasting so much power in the resistor, put two IR LED's in series, this means you can use a MUCH lower value resistor, and output a lot more IR with no increase in current consumption.

Check my tutorial hardware at

 

[gramo]

Instead of wasting so much power in the resistor, put two IR LED's in series, this means you can use a MUCH lower value resistor, and output a lot more IR with no increase in current consumption.

Yet another great idea nigel. Here's where I'm at with the new mod's, sorry about all the meters, just want to be sure

 

[audioguru]

I am glad that you finally dood it.

 

[Nigel Goodwin]

Yet another great idea nigel.

I'd love to take credit for it, but it's common practice in older IR remote controls - where you had four batteries.

Here's where I'm at with the new mod's, sorry about all the meters, just want to be sure

Series resistor seems a bit high?, notice my tutorial only used 4.7 ohm - for IR you're looking for short high current pulses and low duty cycles.

 

[gramo]

Series resistor seems a bit high?, notice my tutorial only used 4.7 ohm - for IR you're looking for short high current pulses and low duty cycles.

I've had issues with this in the past, perhaps a little guidance would help me out,

How do you calculate what resistances to use when PWM signals are used. Obviously as PWM frequency increases, the transistor has less time too turn on and off (not fully driving the IR LED), and the opposite for slower PWM's

 

[Nigel Goodwin]

I've had issues with this in the past, perhaps a little guidance would help me out,

How do you calculate what resistances to use when PWM signals are used. Obviously as PWM frequency increases, the transistor has less time too turn on and off (not fully driving the IR LED), and the opposite for slower PWM's

For IR remote control you pulse the LED at high power but low duty cycle, this maximises range, and keeps average current consumption low.

Speed of switching isn't a problem, as you're only using 38/40KHz, and it's not crucial that the transistor switches rapidly (as it is in switch-mode PSU's).

 

[gramo]

QUOTE=Nigel Goodwin]For IR remote control you pulse the LED at high power but low duty cycle, this maximises range, and keeps average current consumption low.

Speed of switching isn't a problem, as you're only using 38/40KHz, and it's not crucial that the transistor switches rapidly (as it is in switch-mode PSU's).[/QUOTE]

Thanks Nigel, but what lead you to use the 4.7ohm resistor, instead of something higher like you would use if the IR LED was on continuously?

In your tutorial you say;

As the receiver detects 38KHz modulation, we need to pulse the LED's at 38KHz, this can be done by feeding the LED's with a 13uS pulse followed by a 13uS space - in actual fact I decrease the pulse length, and increase the space length (keeping the total length at 26uS) - this reduces the power consumption.

How much did you lower the duty cycle by in real life?

 

[Nigel Goodwin]

QUOTE=Nigel Goodwin]Thanks Nigel, but what lead you to use the 4.7ohm resistor, instead of something higher like you would use if the IR LED was on continuously?

The 4.7 ohm was accurately calculated, after great deliberation

5V supply, 1A maximum required - 5 ohm resistor looks good - completely ignoring the voltage drop across the LED's and transistor!. Obviously, if you were looking for maximum possible range, you could reduce it substantially, but with 4.7 ohm it has a similar range to commercial IR remotes.

How much did you lower the duty cycle by in real life?

I can't even remember now, I think it was something like 40/60?, but the SIRC's system has nice wide gaps in between the pulses anyway, which further reduces dissipation in the LED's.

 

[BOBKA]

Can't you just connect the resistor in series with the IR LED? This way you will control the current with the resistor value. So you will connect it as:
uC_Pin <--> Resistor <--> IR transimiiter <--> GND

If you need 50mA across the IR transmitter (which is simply a diode), assuming that the uC is running at +5V, you will need a 5V/50mA = 100Ohm resistor.

Please correct me if I am conceptually wrong.

 

[audioguru]

Can't you just connect the resistor in series with the IR LED? This way you will control the current with the resistor value. So you will connect it as:
uC_Pin <--> Resistor <--> IR transimiiter <--> GND

If you need 50mA across the IR transmitter (which is simply a diode), assuming that the uC is running at +5V, you will need a 5V/50mA = 100Ohm resistor.

Please correct me if I am conceptually wrong.

Yes you are wrong, very wrong.
1) The absolute max current allowed from a PIC output is only 25mA and then the output voltage will be much less than 5V. So a transistor is needed to drive the LED and its current-limiting resistor.
2) Ohms Law says that current= V/R. The R won't have 5V across it when it is in series with a 1.5V LED. The resistor has only 3.5V across it if the transistor conducts perfectly and only 3.3V across it in practice. So the resistor value must be 3.3V/50mA= 66 ohms or 68 ohms for the nearest value.

 

[BOBKA]

Ok, I figured those out ... but then why can we drive regular LEDs w/o using any of the drivers? Simply connecting them to the PIC works just fine. Does it mean that regular LEDs draw less current?

Also, I guess that's the reason why people use AVR instead of PICs, I remember that someone told me that AVRs can support more currents. Is this true?

 

[Sceadwian]

It's not good for the I/O line when you do this. AVR's and PIC's have roughly the same power output ability on indivial I/O lines.

 

[audioguru]

Ok, I figured those out ... but then why can we drive regular LEDs w/o using any of the drivers? Simply connecting them to the PIC works just fine. Does it mean that regular LEDs draw less current?

The output current of a PIC directly to an LED is from 17ma for a weak one to 35mA for a strong one. The max allowed output current for the PIC is 25mA and the max allowed for most ordinary LEDs is 30mA. Something might blow up without a current-limiting resistor.

 

[BOBKA]

Right, so I guess it is better to put the source follower / emitter follower amplifier in order to increase the drive current capacity I guess this is very important for IR